Question

I have a follow-up question from a problem I posted earlier:

Approximate sin(61π/360):
So, sin(60π/360 + π/360)
= sin(π/6 + (π/360))

Now the equation for linear approximation is f(x+a)=f(a)+f'(a)(x-a). To get the right answer, that means plugging in the numbers like this: sin(61π/360) = sin(π/6) + cos(π/6)*(π/360), where x=π/6 and a=π/360.

So my question is, why does the (x-a) part in the equation equal (π/360)? If it's (x-a), shouldn't it be ((π/6)-(π/360)) which would equal (59π/360)?

Answer 1

<<Now the equation for linear approximation is f(x+a)=f(a)+f'(a)(x-a). To get the right answer, that means plugging in the numbers like this: sin(61π/360) = sin(π/6) + cos(π/6)*(π/360), where x=π/6 and a=π/360. >>
Nope. the Linear approximation is
f(x)=f(a)+f'(a)(x-a)
here, a= 60PI/360 (pi/60) , x=61PI/360, so x-a=PI/360